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Test: Factors & Multiples


20 Questions MCQ Test Mathematics for Class 5: NCERT | Test: Factors & Multiples


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Attempt Test: Factors & Multiples | 20 questions in 20 minutes | Mock test for Class 5 preparation | Free important questions MCQ to study Mathematics for Class 5: NCERT for Class 5 Exam | Download free PDF with solutions
QUESTION: 1

Which of the following statements is correct?

Solution:
QUESTION: 2

Find the number of multiples of 5.

Solution:
QUESTION: 3

How many prime numbers are there "between 1 to 100"?

Solution:

Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 = 25

QUESTION: 4

Which of the following is a perfect square?
Solution:

25 is the perfect square of 5. rest all options are not perfect squares of any natural number.

QUESTION: 5

What is the HCF of two prime numbers?
Solution:
QUESTION: 6

What is the LCM of 36, 75 and 80?
Solution:


LCM = 2×3×5×2×3×5×4 = 3600

QUESTION: 7

Product of two numbers = ___________
Solution:
QUESTION: 8

The sum of the factors of 25 is
Solution:

Factors of = 25 = 52 are 1, 5, 25 which can also be represented as: 52 = (2+1) = 3 (i.e.3 factors) 

QUESTION: 9

A number is divisible by 24 if ___________
Solution:

It is divisible by 8 & 3 because 8 and 3are co-prime

QUESTION: 10

Prime factorization of 512 × 343
Solution:

512×343 =

2×2×2×2×2×2×2×2×2×7×7×7=29 x 73

QUESTION: 11

Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21?
Solution:


Reminder= 40 84040=800 100000+800=100800

QUESTION: 12

182/195 when reduced to the lowest term is 
Solution:

QUESTION: 13

Which of the following pair is co-prime?
Solution:

HCF of 81 and 135 = 27 HCF of 87 and 116 = 29 HCF of 43 and 73 = 1 HCF of 26 and 65 = 13  Hence, 43 and 73 are co-prime

QUESTION: 14

Find the number of factors of 1024
Solution:

1024=210=2n
⇒ (n+1) = (10+1) = 11.
Where (n+1) =  no. of factors. Here factors will be 1, 2, 4, 8, 16?????.1024.

QUESTION: 15

If the 8-digit number 136y5785 is divisible by 15, then find the least possible value of
Solution:

136Y5785

For 15, No. should be divisible by 5, 3
No. is already divisible by 5 1 + 3 + 6 + 5 + 7 + 8 + 5 + y = 35 + y

∴y=1

QUESTION: 16

Find the smallest number which when divided by 25, 40 and 60 leaves remainder 7 in each case?
Solution:


5×4×5×2×3=600
Required numbers is 600+7=607

QUESTION: 17

x, y, z, w are four odd natural numbers. Let u=x2+y2+z2+w2. Consider following statements

I. U is always divisible by 4
II. U is never divisible by 8

Q. Which of the above statements) is /are true?
Solution:
QUESTION: 18

2211/5025 , which reduced to its simplest form. By dividing the numerator and denominator
Solution:

QUESTION: 19

A boy saves Rs. 4.65 daily. Find the least number of days in which he will be to save an exact number (whole number) of rupees.
Solution:

Rs. 4.65 = 465 paise

The exact number of rupees will be a multiple of 100

So, we have to fine the L.C.M. of 465 and 100

100 = 2*2*5*5

465 = 3*5*31

L.C.M. of 465 and 100 = 2*2*3*5*5*31

= 9300

So, the boy saves 9300 paise

Hence, least number of days required to save 9300 paise = 9300/465

= 20 days

QUESTION: 20

H.C.F of two numbers is 16, then their LCM is
Solution:

H.C.F is always a factor of L.C.M. In this case 16 Divides 192 only.

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